 Given equation:

$$Ca(OH)_{2} + 2NH_{4}Cl \rightarrow CaCl_{2} + 2NH_{3} + 2H_{2}O .$$

Molar mass of $$NH_{4}Cl = 53.5 g/mole$$

Molar mass of $$CaCl_{2} = 111 g/mole$$

Molar mass of Ammonia $$(NH_{3}) = 17 g/mole$$

1. The Mass Of Calcium Chloride Formed

Two moles of $$NH_{4}Cl react to give one mole of CaCl_{2}$$

Two moles of $$NH_{4}Cl will give$$ $$\Rightarrow \frac{0.1 * 1}{2}$$ $$\Rightarrow 0.05 moles of CaCl_{2}$$ $$Mass of CaCl_{2} = 0.05 * 111$$ $$\Rightarrow 5.55 grams of CaCl_{2}$$

Therefore,the mass of Calcium Chloride formed is 5.55 gms

1. The Volume, At S.T.P Of Ammonia Liberated.

Two moles of $$NH_{4}Cl gives one mole of NH_{3} gas$$ $$0.1 moles of NH_{4}Cl will react to give$$ $$\Rightarrow \frac{0.1}{2} = 0.05 moles of NH_{3} gas$$

One mole of Ammonia $$(NH_{3}) at STP = 2.4 L$$ $$0.05 mole of Ammonia (NH_{3}) at STP = 22.4 * 0.5$$ $$\Rightarrow 1.12 L of NH_{3} gas$$

Therefore, the volume, at S.T.P of Ammonia gas liberated is 1.12 L

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